A point charge q1 is at the center of a sphere of radius 20 cm. Another point charge q2 = 5 nC is located at a distance r = 30 cm from the center of the sphere. If the net flux through the surface of the sphere is 800 N.m2 /C, find q

Respuesta :

Answer:

[tex]q_1 =7.08*10^{-9}C.[/tex]

Explanation:

Gauss's Law says that the electric flux [tex]\Phi_E[/tex] through a closed surface is directly proportional to the charge [tex]Q_{enc}[/tex] inside it. More precisely,

[tex]$\Phi_E=\oint_S E\cdot dA = \dfrac{Q_{enc}}{\epsilon_0}. $[/tex]

This means what is outside this closed surface [tex]S[/tex] does not contribute to the flux through it because field lines that go in must come out, resulting a zero flux from an external charge.

In our context, this means the charge [tex]q_2[/tex] which is outside the sphere will have zero flux through the surface; therefore, Gauss's law will only be concerned with charge [tex]q_1[/tex] which is inside the sphere; Hence,

[tex]$\Phi_E=\oint_S E\cdot dA = \dfrac{q_1}{\epsilon_0} = 800 N\cdot m^2/C. $[/tex]

Solving for [tex]q_1[/tex] gives

[tex]$ q_1= (800 N\cdot m^2/C)\epsilon_0, $[/tex]

[tex]$ q_1= (800 N\cdot m^2/C)*(8.85*10^{-12}C^2/N\cdot m^2) $[/tex]

[tex]\boxed{q_1 =7.08*10^{-9}C. }[/tex]

which is the charge inside the sphere.